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Falling elevator onto a spring.

  1. Jul 6, 2015 #1
    1. The problem statement, all variables and given/known data
    The cable of the 2000 kg elevator cab in the figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 2.8 m above a spring of spring constant k = 0.30 MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 2.4 kN opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance x that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance (above the point of maximum compression) that the cab will bounce back up the shaft. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)

    http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c08/fig08_59.jpg

    2. Relevant equations
    20150706_205005_zpsglzzvreg.jpg

    3. The attempt at a solution
    I got part a, but part b is where im stuck. i got an answer of .0573m im not sure how to find distance without using the spring equation.
    20150706_204914_zps9hqcp0i4.jpg
     
  2. jcsd
  3. Jul 6, 2015 #2

    Nathanael

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    x=(mg-Ff)/k is when the acceleration is zero, but you want the velocity to be zero not acceleration.

    If the spring is compressed by an amount x, then how much work from gravity and friction is done? And if the velocity is zero, then what is the spring-potential-energy?
     
  4. Jul 6, 2015 #3
    x amount would be w=1/2 kx2 + 48160J.
    If velocity is zero then spring potential energy is at the maximum.
    Edit
    work done = 17200d
    max being = 48160J
     
    Last edited: Jul 6, 2015
  5. Jul 6, 2015 #4

    Nathanael

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    Gravity and friction still do work while the spring is being compressed, so it should depend on x.
     
  6. Jul 6, 2015 #5
    would work be (19600+2400)x? because the spring is working against friction? But so is the gravity force.
     
  7. Jul 6, 2015 #6

    Nathanael

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    I'm not sure what you mean by 'the spring works against friction.' Gravity pulls it down (positive work since it's falling) and the spring and friction push it up (negative work).

    You had the work done by gravity and friction almost correct before when you put (mg-Ff)d but "d" is not just the distance above the spring, it also depends on how much the spring is compressed.
     
  8. Jul 6, 2015 #7
    work=Fd=17200(2.8)=48160J.
    48160=1/2 kx2. but when I solve for x i get .566 its wrong
     
  9. Jul 6, 2015 #8

    Nathanael

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    d is not 2.8, d depends on x
     
  10. Jul 6, 2015 #9
    d=2.8m was given. im using it for the distance to find work down by the falling elevator before it touches the spring. when the spring is fully compressed its work should be the same but opposite.
     
  11. Jul 6, 2015 #10

    Nathanael

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    2.8m is given as the distance between the spring and the bottom of the elevator... But the elevator falls farther than that doesn't it? If the elevator only fell 2.8m then the spring wouldn't be compressed.
     
  12. Jul 6, 2015 #11
    That is the distance im trying to find. I was thinking if the energy of the elevator hits the spring with 48160J the spring must compress and equal that energy when it comes to a stop at distance x in the formula W= 1/2 kx2
     
  13. Jul 6, 2015 #12

    SammyS

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    Adding to what Nathanael posted ...

    You have found the work done by gravity and friction in falling prior to contact with the spring. How much work do gravity and friction do over the distance, x, when the spring is being compressed?

    (I would have thought that you would have these spring problems licked by now.)
     
  14. Jul 6, 2015 #13
    these problems are licking me.
    when the spring is being compressed it does a total of -48160J? because its oppose of the KE

    edit; or would it be w=17200x where x is the distance the spring is compressed.
     
  15. Jul 6, 2015 #14

    Nathanael

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    48160J is the work done on the elevator before the spring is compressed... But as the spring compresses, gravity and friction still act on the elevator (doing additional work).
    Your mistake is basically the same as your mistake in post #4 of your previous problem.
     
  16. Jul 6, 2015 #15

    SammyS

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    BINGO!

    It's the 17200x which you have been neglecting and Nathanael has been hinting at strongly.
     
  17. Jul 6, 2015 #16
    .30M N/M = 300000NM?
     
  18. Jul 6, 2015 #17

    Nathanael

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    Yes, 0.3 Mega N/m is 300,000 N/m
     
  19. Jul 6, 2015 #18
    but then i get 2.8m again
    48160 +17200x = W, solve for x its 2.8m
     
  20. Jul 6, 2015 #19

    Nathanael

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    By "W" you mean the work done by the spring, right? You should get a quadratic equation.
     
  21. Jul 6, 2015 #20
    W=1/2 k (48160+17200x)x im really confused.
     
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